In a redox reaction:H2 + F2 =2HF, the number of electrons involved in the conversion is 1. 0 2. 2 3. 4 4. 5?

Redox reaction.......

Redox reactions are sometimes described by the hypothetical "transfer" of electrons. We write half-reactions using the gain or loss of electrons, but with the understanding that these transfers are hypothetical.
Oxidation - loss of electrons
Reduction - gain of electrons

A more comprehensive defintion:
Oxidation - increase in oxidation state
Reduction - decrease in oxidation state

0 .......... 0 ........... +1-1 ...... oxidation states
H2(g) + F2(g) --> 2HF(g)
There is a hypothetical "transfer" of 1 electron per HF molecule, or 2 electrons for the 2 HF molecules shown in the balanced chemical equation. The answer is 2 electrons. The rationale is shown below.

H2 + 2F- --> 2HF + 2e-
F2 + 2H+ 2e- --> 2HF
------------- ---------------- --------------
H2 + F2 + 2H+ + 2F- --> 4HF
simplify
H2 + F2 --> 2HF

H₂ + F₂ → 2HF

The oxidation number of H increases from 0 (in H₂) to +1 (in HF).
For each H atom, 1 electron is lost as the change in oxidation number is +1.
According to the equation, there are 2 H atoms. Hence, 2 electrons are lost.

On the other hand, the oxidation number of F decrease from 0 (in H₂) to -1 (in HF).
For each F atom, 1 electron is gained as the change in oxidation number is -1.
According to the equation, there are 2 F atoms. Hence, 2 electrons are gained.

...... The answer is : 2. 2

4 electrons. 1 electron from each atom of hydrogen and fluorine react to form a bond. 4 atoms give 2 bonds